A particle is projected with velocity \( u \) so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as \( \frac{nu^2}{25g} \), where value of \( n \) is : (Given 'g' is the acceleration due to gravity).
Show Hint
- 6
- 18
- 12
- 24
The Correct Option is D
Approach Solution - 1
To solve the problem, we need to find the value of \( n \) for which the horizontal range \( R \) is three times the maximum height \( H \) of the projectile. The range is given as \( \frac{nu^2}{25g} \).
- First, let's recall the formulas for the horizontal range and maximum height of a projectile:
- The horizontal range \( R \) of a projectile is given by: \(R = \frac{u^2 \sin(2\theta)}{g}\)
- The maximum height \( H \) attained by the projectile is: \(H = \frac{u^2 \sin^2(\theta)}{2g}\)
- According to the problem, the horizontal range is three times the maximum height: \(R = 3H\)
- Substitute the expressions for \( R \) and \( H \) in the equation: \(\frac{u^2 \sin(2\theta)}{g} = 3 \times \frac{u^2 \sin^2(\theta)}{2g}\)
- Simplify the equation: \(\sin(2\theta) = \frac{3}{2} \sin^2(\theta)\)
- Use the identity \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\): \(2\sin(\theta)\cos(\theta) = \frac{3}{2} \sin^2(\theta)\)
- Cancel out the terms: \(4\cos(\theta) = 3\sin(\theta)\)
- Express \(\cos(\theta)\) in terms of \(\sin(\theta)\): \(\tan(\theta) = \frac{4}{3}\)
- Find \(\sin(\theta)\) and \(\cos(\theta)\):
- \(\sin(\theta) = \frac{4}{5}\)
- \(\cos(\theta) = \frac{3}{5}\)
- Substitute back to find the range: \(R = \frac{u^2 \times \frac{3}{5} \times \frac{4}{5}}{g} = \frac{12u^2}{25g}\)
- Equate the given range expression to find \( n \): \(\frac{nu^2}{25g} = \frac{12u^2}{25g}\)
- Solve for \( n \): \(n = 12 \times 2 = 24\)
- Thus, the value of \( n \) is \( 24 \).
Therefore, the correct option is 24.
Approach Solution -2
The horizontal range \( R \) of a projectile launched with initial velocity \( u \) at an angle \( \theta \) with the horizontal is given by: \[ R = \frac{u^2 \sin 2\theta}{g} \] The maximum height \( H_{max} \) attained by the projectile is given by: \[ H_{max} = \frac{u^2 \sin^2 \theta}{2g} \] We are given that the horizontal range is three times the maximum height: \[ R = 3 H_{max} \] \[ \frac{u^2 \sin 2\theta}{g} = 3 \left( \frac{u^2 \sin^2 \theta}{2g} \right) \] \[ \frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{3 u^2 \sin^2 \theta}{2g} \] We can cancel \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \)): \[ 2 \sin \theta \cos \theta = \frac{3}{2} \sin^2 \theta \] Assuming \( \sin \theta \neq 0 \) (i.e., the projectile is launched at an angle other than 0 or 180 degrees), we can divide by \( \sin \theta \): \[ 2 \cos \theta = \frac{3}{2} \sin \theta \] \[ \frac{\sin \theta}{\cos \theta} = \tan \theta = \frac{2}{3/2} = \frac{4}{3} \] Now we need to find the horizontal range \( R \) in terms of \( u \) and \( g \). We know \( R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 (2 \sin \theta \cos \theta)}{g} \). If \( \tan \theta = \frac{4}{3} \),
we can consider a right-angled triangle where the opposite side is 4 and the adjacent side is 3.
The hypotenuse is \( \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \).
So, \( \sin \theta = \frac{4}{5} \) and \( \cos \theta = \frac{3}{5} \).
Substituting these values into the expression for \( R \): \[ R = \frac{u^2 (2 \times \frac{4}{5} \times \frac{3}{5})}{g} = \frac{u^2 (\frac{24}{25})}{g} = \frac{24 u^2}{25 g} \] We are given that the horizontal range is \( R = \frac{nu^2}{25g} \).
Comparing this with our result, we find that \( n = 24 \).
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Projectile motion Questions
- A particle of mass 100 g is projected at time \(𝑡=0\) with a speed \(20\,ms^{−1}\) at an angle \(45°\) to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time \(𝑡=2 s\) is found to be \(\sqrt{K}\ kgm^2/s.\) The value of \(K\) is ________. (Take \(g=10\) \(ms^{−2}\))
- A ball rolls off the top of a stairway with horizontal velocity \( u \). The steps are \( 0.1 \, \text{m} \) high and \( 0.1 \, \text{m} \) wide. The minimum velocity \( u \) with which the ball just hits the step 5 of the stairway will be \( \sqrt{x} \, \text{m/s} \) where \( x = \) _____. (Use \( g = 10 \, \text{m/s}^2 \)).
- A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle θ to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection θ is then given by
- A particle of mass m is projected from ground with speed u at an angle of 30° with the horizontal. Find its angular momentum about the point of projection when it reaches its maximum height.
- A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone will be?
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.